Question

The equation of the common tangent with positive slope to the parabola $y^{2}=8\sqrt{3x}$ and the hyperbola $4x^2 - y^2 = 4$ is

• A. $y=\sqrt{6}\,x+\sqrt{2}$
• B. $y=\sqrt{6}\,x-\sqrt{2}$
• C. $y=\sqrt{3}\,x+\sqrt{2}$
• D. $y=\sqrt{3}\,x-\sqrt{2}$

Answer 1

The Correct Option is A

Equation of tangent in slope form of parabola $y^{2}=8 \sqrt{3} x$ is
$y=m x+c\,\,\,...(i)$
where,$c =\frac{a}{m} $
$\therefore c =\frac{2 \sqrt{3}}{m}\,\,\,...(ii)$
Also, tangent to the hyperbola
$4 x^{2}-y^{2}=4$
or $\frac{x^{2}}{1}-\frac{y^{2}}{4}=1$ is
$c^{2}=a^{2} m^{2}-b^{2} $
$c^{2}=1 m^{2}-4$
$\Rightarrow \left(\frac{2 \sqrt{3}}{m}\right)^{2}=m^{2}-4 \,\,\,$ [from E (ii)]
$\Rightarrow \frac{12}{m^{2}}=m^{2}-4 $
$ \Rightarrow m^{4}-4 m^{2}-12=0 $
$ \Rightarrow m^{4}-6 m^{2}+2 m^{2}-12=0 $
$ \Rightarrow m^{2}\left(m^{2}-6\right)+2\left(m^{2}-6\right)=0 $
$\Rightarrow \left(m^{2}+2\right)\left(m^{2}-6\right)=0 $
$ \Rightarrow m^{2}-6=0 $ and $m^{2}+2 \neq 0 $
$ \Rightarrow m^{2}=6 $
$ \Rightarrow m=\pm \sqrt{6}$
i.e., $m=\sqrt{6}$ as $m$ is positive slope.
$\therefore$ From E (i),
$y=\sqrt{6} x+\frac{2 \sqrt{3}}{\sqrt{6}}$
$\Rightarrow y=\sqrt{6} x+\sqrt{2}$